Explanation

(a)
Let $I = \int_{0}^{\pi/4}{\log{(1 + \tan{x)dx\ }}\ldots.(i)}$
$\Rightarrow \ I = \int_{0}^{\pi/4}{\log{\left\lbrack 1 + \tan\left( \frac{\pi}{4} - x \right) \right\rbrack dx}}$
$= \int_{0}^{\pi/4}{\log{\left\lbrack 1 + \frac{1 - \tan x}{1 + \tan x} \right\rbrack dx}}$
$= \int_{0}^{\pi/4}{\log{2dx - \int_{0}^{\pi/4}{\log{(1 + \tan{x)dx}}}}}$
$\Rightarrow \ I = \log{2\lbrack x\rbrack_{0}^{\pi/4}} - I$
$\Rightarrow \ 2I = \frac{\pi}{4}\log_{e}{2\ \Rightarrow \ I = \frac{\pi}{8}\log_{e}2}$